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Question
In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that
(i) R is the mid-point of BC, and
(ii) PR = `(1)/(2)"DB"`.
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Solution
(i) Join B and D. Suppose AC and BD cut at O. Then,
OC = `(1)/(2)"AC"`
Now,
CQ = `(1)/(4)"AC"`
⇒ CQ = `(1)/(2)"OC"`
In ΔDCO, P and Q are the mid-points of DC and OC respectively.
∴ PQ || DO
Also, in ΔCOB, Q is the mid-point of OC and PQ || OB
Therefore, R is the mid-point of BC, R being PQ produced.
(ii) In ΔBCD, P and R are the mid-points of DC and BC respectively.
Also PR || BD
Therefore, PR = `(1)/(2)"BD"`.
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